@Aditri
Mol of rod passing through the center= 1/12 ML²
Mol of rod passing through one side= 1/3 ML²
AP= √3/2 L
Mol of BC through the prependicular A
= 1/12 ML² + M(√3/2 L)²
= 1/12 ML² + √3/4 ML²
= 5/6 ML²
By the principle of superposition
= 1/3 ML² + 1/3 ML² + 5/6 ML²
= 9/6ML²
= 3/2 ML²
radius of gyration
k = √3/2 ML²/ 3M
= L/√2
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